Solve Laplace's equation by separation of variables in cylindrical coordinates, assuming there is no dependence on z (cylindrical symmetry). [Make

sure you find all solutions to the radial equation; in particular, your result must accommodate the case of an infinite line charge, for which (of course) we already know the answer.]

Write the expression for Laplace in terms of cylindrical coordinates.

$\frac{\mathbf{1}}{\mathbf{s}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{s}}\left(s\frac{\partial V}{\partial s}\right)\mathbf{+}\frac{\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{V}}{\mathbf{\partial }{\mathbf{\varphi }}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{V}}{\mathbf{\partial }{\mathbf{z}}^{\mathbf{2}}}\mathbf{=}\mathbf{0}$ ……. (1)

Given that there is no independent of z. Then the Laplace equation becomes,

$\frac{\mathbf{1}}{\mathbf{S}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{s}}\left(s\frac{\partial V}{\partial s}\right)\mathbf{+}\frac{\mathbf{1}}{{\mathbf{S}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{V}}{\mathbf{\partial }{\mathbf{\varphi }}^{\mathbf{2}}}\mathbf{=}\mathbf{0}$ …… (2)

The solution for the above equation in the following form,

$V\left(S,\varphi \right)=S\left(s\right)\Phi \left(\varphi \right)$ …….. (3)

Here, s and $\Phi$are two functions.

Substitute $S\left(s\right)\Phi \left(\phi \right)$for $V\left(S,\phi \right)$in equation (2).

$$\begin{gathered} \frac{1}{s}\frac{\partial }{\partial s}\left(s\frac{\partial }{\partial s}\left(S\left(s\right)\Phi \left(\varphi \right)\right)\right)+\frac{1}{S^2}\frac{{\partial }^1}{\partial {\varphi }^2}\left(S\left(s\right)\Phi \left(\varphi \right)\right)=0 \\ \frac{1}{s}\Phi \frac{d}{ds}\left(s\frac{dS}{ds}\right)+\frac{1}{s^2}S\frac{d^2\Phi }{d{\varphi }^2}=0 \end{gathered}$$

Multiply to above equation by $\frac{s^2}{S\Phi }$on both sides.

$\frac{s}{S}\frac{d}{ds}\left(s\frac{dS}{ds}\right)+\frac{1}{\Phi }\frac{d^2\Phi }{d{\varphi }^2}=0$ …….. (4)

Substitute $\frac{s}{S}\frac{d}{ds}\left(s\frac{dS}{ds}\right)=C_1,\frac{1}{\Phi }\frac{d^2\Phi }{d{\varphi }^2}=C_2$in equation (4)

Therefore,

$C_1+C_2=0$

To solve above equation assume that $C_2=-K^2$

$\frac{1}{\Phi }\frac{d^2\Phi }{d{\varphi }^2}=C_2$

Substitute $C_2=-K^2$in above equation

$$\begin{gathered} \frac{1}{\Phi }\frac{d^2\Phi }{d{\varphi }^2}=-K^2 \\ \frac{d^2\Phi }{d{\varphi }^2}+K^2\Phi =0 \end{gathered}$$ ......(5)

The solution for the above equation is,

$\Phi =A\cos K\varphi +B\sin K\varphi$

Here, A,B are constant.

Thus the value of $C_1$is,

$C_1=K^2$.

Substitute $\frac{d}{S}\frac{d}{ds}\left(s\frac{dS}{ds}\right)$for $C_1$.

$$\begin{gathered} s\frac{d}{ds}\left(sn{s}^{n-1}\right)=K^2{s}^n \\ ns\frac{d}{ds}{s}^n=K^2{s}^n \\ {n}^{2}s{s}^{n-1}=K^2{s}^n \\ {n}^2{s}^n=K^2{s}^n \end{gathered}$$

Substitute $s^n$for S to solve the above equation

$$\begin{gathered} s\frac{d}{ds}\left(sn{s}^{n-1}\right)=K^2{s}^n \\ ns\frac{d}{ds}{s}^n=K^2{s}^n \\ {n}^{2}s{s}^{n-1}=K^2{s}^n \\ {n}^2{s}^n=K^2{s}^n \end{gathered}$$

Solve the above equation for $n$, here $\left(n=\pm K\right)$.

Solve the equation,

$$\begin{gathered} s\frac{d}{ds}\left(s\frac{dS}{ds}\right)=K^{2}S \\ {s}^2\frac{d^{2}S}{d{s}^2}+\frac{d^{2}S}{d{s}^2}-K^{2}S=0 \end{gathered}$$ ......(6)

This is Euler’s second order homogenous differential equation.

The solution for the above equation is

$S\left(s\right)=C{s}^k+D{s}^{-k}$

Here, $C$and $D$are constant.

Unless$K=0$, there are only one solution of second order $S$is equal to the constant.

Thus, keep $K=0$separately.

Substitute 0 for K, in differential equation (6).

$$\begin{gathered} s\frac{d}{ds}\left(s\frac{dS}{ds}\right)=0 \\ s\frac{dS}{ds}=Cons\tan t \\ \frac{dS}{ds}=\frac{E}{s} \\ S=EIns+F \end{gathered}$$

Here, E and F are constant.

Substitute 0 for K in the differential equation (5)

$\frac{1}{\Phi }\frac{d^2\Phi }{d\varphi }=0$

Integrate twice the above equation,

$$\begin{gathered} \frac{d\Phi }{d\varphi }=G \\ d\Phi =Gd\varphi \\ \Phi =G\varphi +H \end{gathered}$$

Therefore,

Write the general solution in cylindrical symmetry.

$V\left(S,\varphi \right)=\sum _{K=1}^{\infty }\left[\begin{array}{c}{S}^k\left(a_k\cos K\varphi +b_k\sin K\varphi \right)\\ +s^{-K}\left(C_k\cos K\varphi +d_K\sin K\varphi \right)\end{array}\right]+\left(EIns+F\right)+\left(F\varphi +H\right)$ ......(7)

The term $G\varphi$is unacceptable term.

Therefore, neglect that term.

Thus equation (7) becomes,

$$\begin{gathered} V\left(S,\varphi \right)=\sum _{K=1}^{\infty }\left[\begin{array}{c}{S}^k\left(s_K\cos K\varphi +b_K\sin K\varphi \right)\\ +s^{-K}\left(c_k\cos K\varphi +d_K\sin K\varphi \right)\end{array}\right]+a_0{b}_{0}Ins \\ =a_0+b_{0}Ins+\sum _{K=1}^{\infty }\left[\begin{array}{c}{S}^k\left(s_K\cos K\varphi +b_K\sin K\varphi \right)\\ +s^{-K}\left(c_k\cos K\varphi +d_K\sin K\varphi \right)\end{array}\right] \end{gathered}$$

Here, $b_0=E$and $a_0=F+H$are constant.

Hence, solution for the Laplace equation in cylindrical coordinates is,

$V\left(S,\varphi \right)=a_0+b_{0}Ins+\sum _{K=1}^{\infty }\left[\begin{array}{c}{S}^k\left(a_K\cos K\varphi +b_K\sin K\varphi \right)\\ +s^{-K}\left(c_k\cos K\varphi +d_K\sin K\varphi \right)\end{array}\right]$.